A Spring-y Problem
Most recent answer: 10/22/2007
Q:
How do I work out the following problem??
"A spring system is constructed by connecting two springs in parrallel. The spring constant of each spring is 120nm(to the minus 1) Calculate the mass that is needed to be suspended from the system if the extension is to be 20cm. State any assumptions you have made."
- Lucy (age 16)
Stourbridge
- Lucy (age 16)
Stourbridge
A:
Lucy -
First of all, to make this simpler I’m first going to explain this as if there’s only 1 spring, rather than 2 connected in parallel...
The first most important thing to do when solving a problem like this is to draw a picture! (Even if you can picture it in your head, having a picture on paper is gauranteed to make your life easier in the long run!)
The big thing that is going to make solving this problem a lot simpler is that we are going to assume that nothing is moving. If the mass were bouncing up and down on the spring, for instance, this problem would get a lot more interesting. ;)
Since the mass isn’t moving, we know that the net force on it has to be zero. (This is one of the basic laws of physics: F=ma. An object that’s not moving isn’t accelerating, so a=0. So F = m*a = m*0 = 0. Note: this doesn’t mean that there aren’t any forces acting on the object - just that they add up to a total of zero.) If the net force were anything other than zero, the mass would be speeding up or slowing down - not sitting totally still.
So we know that all the forces on the mass have to add up to zero. Now all we need is a list of the forces. These are the force of gravity pulling down on the mass and the force of the spring pulling up. Since the two forces are pulling in opposite directions, we’re going to give them opposite signs (making Fspring positive and Fgravity negative):
Fspring - Fgravity = 0
We’re going to make another assumption here and say that we’re operating in an "inertial reference frame". For our purposes, this is equivalent to saying that the whole spring and mass set-up isn’t speeding up or slowing down - that is, the spring isn’t hanging inside of an accelerating car. Technically, the earth isn’t a perfect inertial reference frame, since the earth is spinning and circling the sun (among other interesting things), but for our purposes the effects of these movements are so small we can just ignore them.
So, sticking with the above equation we have Fspring - Fgravity = 0 or:
kx - mg = 0
kx = mg
(Here comes assumption number 3! We’re assuming that these are "ideal" springs - that is, they don’t get stressed or lose their springiness over time like real springs do. We’re assuming that the force from the spring follows the neat tidy equation Fspring = kx, where k is the spring constant, and x is the extra distance over which the spring is stretched.)
Let’s plug in for gravity (9.81 m/s2and the distance x (20 cm = 0.2 m) and work out m:
0.2k = 9.81m
m = 0.0204k
To find the mass (in kg), all you have to do is plug in the spring constant (in units of kg/s2).
But wait! Your problem had 2 springs:
But really, what’s the difference? Well, now we have 2 forces pulling upwards (positive) and one pulling downwards (negative):
Fspring 1 + Fspring 2 - Fgravity = 0
kx + kx - mg = 0
2kx = mg
You could solve it from here just like above - by plugging in your numbers. But look at those equations!
1 spring: kx = mg
2 springs: 2kx = mg
It turns out that if we kept adding more springs, we’d get the same sort of equations:
3 springs: 3kx = mg
4 springs: 4kx = mg
5 springs: 5kx = mg
etc...
Really, using several springs in parallel is just like having one big spring. And to figure out the (imaginary) spring constant of that one big spring, all you have to do is add the spring constants of all the smaller springs up.
Having said all that, how would you solve a problem with 5 springs all rigged up in parallel, with spring constants of 1 kg/s2, 2 kg/s2, 3 kg/s2, 4 kg/s2, 5 kg/s2 respectively? (Hint: you don’t have to work out the forces of all 5 springs separately!)
-Tamara
First of all, to make this simpler I’m first going to explain this as if there’s only 1 spring, rather than 2 connected in parallel...
The first most important thing to do when solving a problem like this is to draw a picture! (Even if you can picture it in your head, having a picture on paper is gauranteed to make your life easier in the long run!)
The big thing that is going to make solving this problem a lot simpler is that we are going to assume that nothing is moving. If the mass were bouncing up and down on the spring, for instance, this problem would get a lot more interesting. ;)
Since the mass isn’t moving, we know that the net force on it has to be zero. (This is one of the basic laws of physics: F=ma. An object that’s not moving isn’t accelerating, so a=0. So F = m*a = m*0 = 0. Note: this doesn’t mean that there aren’t any forces acting on the object - just that they add up to a total of zero.) If the net force were anything other than zero, the mass would be speeding up or slowing down - not sitting totally still.
So we know that all the forces on the mass have to add up to zero. Now all we need is a list of the forces. These are the force of gravity pulling down on the mass and the force of the spring pulling up. Since the two forces are pulling in opposite directions, we’re going to give them opposite signs (making Fspring positive and Fgravity negative):
Fspring - Fgravity = 0
We’re going to make another assumption here and say that we’re operating in an "inertial reference frame". For our purposes, this is equivalent to saying that the whole spring and mass set-up isn’t speeding up or slowing down - that is, the spring isn’t hanging inside of an accelerating car. Technically, the earth isn’t a perfect inertial reference frame, since the earth is spinning and circling the sun (among other interesting things), but for our purposes the effects of these movements are so small we can just ignore them.
So, sticking with the above equation we have Fspring - Fgravity = 0 or:
kx - mg = 0
kx = mg
(Here comes assumption number 3! We’re assuming that these are "ideal" springs - that is, they don’t get stressed or lose their springiness over time like real springs do. We’re assuming that the force from the spring follows the neat tidy equation Fspring = kx, where k is the spring constant, and x is the extra distance over which the spring is stretched.)
Let’s plug in for gravity (9.81 m/s2and the distance x (20 cm = 0.2 m) and work out m:
0.2k = 9.81m
m = 0.0204k
To find the mass (in kg), all you have to do is plug in the spring constant (in units of kg/s2).
But wait! Your problem had 2 springs:
But really, what’s the difference? Well, now we have 2 forces pulling upwards (positive) and one pulling downwards (negative):
Fspring 1 + Fspring 2 - Fgravity = 0
kx + kx - mg = 0
2kx = mg
You could solve it from here just like above - by plugging in your numbers. But look at those equations!
1 spring: kx = mg
2 springs: 2kx = mg
It turns out that if we kept adding more springs, we’d get the same sort of equations:
3 springs: 3kx = mg
4 springs: 4kx = mg
5 springs: 5kx = mg
etc...
Really, using several springs in parallel is just like having one big spring. And to figure out the (imaginary) spring constant of that one big spring, all you have to do is add the spring constants of all the smaller springs up.
Having said all that, how would you solve a problem with 5 springs all rigged up in parallel, with spring constants of 1 kg/s2, 2 kg/s2, 3 kg/s2, 4 kg/s2, 5 kg/s2 respectively? (Hint: you don’t have to work out the forces of all 5 springs separately!)
-Tamara
(published on 10/22/2007)